Ideal gas equation and Related gas laws Questions in English

(B) The graph shows a linear relationship between volume $(V)$ and temperature in degrees Celsius $(t)$.
According to Charles's Law,the volume of a fixed mass of gas is directly proportional to its absolute temperature ($T$ in Kelvin).
The relationship is given by: $V_t = V_0(1 + \frac{t}{273.15})$
This can be rewritten as: $V_t = V_0 + (\frac{V_0}{273.15})t$
Comparing this with the linear equation $Y = mx + C$,where $Y = V_t$,$x = t$,$m = \frac{V_0}{273.15}$,and $C = V_0$.
This linear plot of $V$ versus $t$ $(^{\circ}C)$ is characteristic of Charles's Law.
Therefore,the correct option is $(B)$.

(A) According to the ideal gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given: $V_1 = 2 \ L$,$P_2 = 2P_1$,and $T_2 = 2T_1$.
Substituting the values into the equation:
$\frac{P_1 \times 2 \ L}{T_1} = \frac{(2P_1) \times V_2}{(2T_1)}$.
Solving for $V_2$:
$V_2 = \frac{P_1 \times 2 \ L \times 2T_1}{T_1 \times 2P_1} = 2 \ L$.
Thus,the volume remains $2 \ L$.

(D) According to Boyle's Law,at constant temperature,$P_1V_1 = P_2V_2$.
Let the volume of each vessel be $V$. The total volume after joining the vessels becomes $2V$.
Using the principle of conservation of moles (or Dalton's Law of partial pressures),the total pressure $P_{mix}$ is given by:
$P_{mix} = \frac{P_1V_1 + P_2V_2}{V_{total}}$
$P_{mix} = \frac{100 \times V + 400 \times V}{V + V} = \frac{500V}{2V} = 250 \ mm$.

(A) According to Boyle's Law,for a fixed amount of gas at constant temperature,$P_1V_1 = P_2V_2$.
Given: $P_1 = 10^4 \ Pa$,$V_1 = 100 \ cc$,$P_2 = 10^5 \ Pa$.
Substituting the values: $10^4 \times 100 = 10^5 \times V_2$.
$V_2 = \frac{10^4 \times 100}{10^5} = \frac{10^6}{10^5} = 10 \ cc$.

(B) The density $(d)$ of an ideal gas is given by the formula $d = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature in Kelvin.
For a given gas like Neon,$M$ and $R$ are constant,so $d \propto \frac{P}{T}$.
To maximize density,we need the highest pressure $(P)$ and the lowest temperature $(T)$.
Comparing the options:
$A$: $STP$ $(0\,^{\circ}C, 1 \ atm)$ $\rightarrow T = 273 \ K, P = 1 \ atm$ $\rightarrow \frac{P}{T} = \frac{1}{273} \approx 0.00366$
$B$: $0\,^{\circ}C, 2 \ atm$ $\rightarrow T = 273 \ K, P = 2 \ atm$ $\rightarrow \frac{P}{T} = \frac{2}{273} \approx 0.00732$
$C$: $273\,^{\circ}C, 1 \ atm$ $\rightarrow T = 546 \ K, P = 1 \ atm$ $\rightarrow \frac{P}{T} = \frac{1}{546} \approx 0.00183$
$D$: $273\,^{\circ}C, 2 \ atm$ $\rightarrow T = 546 \ K, P = 2 \ atm$ $\rightarrow \frac{P}{T} = \frac{2}{546} \approx 0.00366$
The ratio $\frac{P}{T}$ is highest for option $B$.

(C) The ideal gas equation is $PV = nRT$.
Since $n = \frac{W}{M}$,where $W$ is mass and $M$ is molar mass,we have $PV = \frac{W}{M} RT$.
Rearranging for density $d = \frac{W}{V}$,we get $P = \frac{dRT}{M}$,which implies $d = \frac{PM}{RT}$.
Given: $P = 5.00\ atm$,$M(N_2) = 28\ g/mol$,$T = 227 + 273 = 500\ K$,and $R = 0.082\ L\ atm\ K^{-1}\ mol^{-1}$.
Substituting the values: $d = \frac{5.00 \times 28}{0.082 \times 500} = \frac{140}{41} \approx 3.41\ g/L$.
Note: The density is $3.41\ g/L$. Since $1\ L = 1000\ mL$,the density in $g/mL$ is $0.00341\ g/mL$. Given the options,the numerical value $3.41$ is expected.

(A) Using the combined gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$,we can express the ratio of volumes as $\frac{V_2}{V_1} = \frac{P_1 T_2}{P_2 T_1}$.
Given initial conditions: $T_1 = 15 + 273 = 288 \ K$ and $P_1 = 1.5 \ bar$.
Given final conditions: $T_2 = 25 + 273 = 298 \ K$ and $P_2 = 1.0 \ bar$.
Substituting the values: $\frac{V_2}{V_1} = \frac{1.5 \times 298}{1.0 \times 288} = \frac{447}{288} \approx 1.552$.
Rounding to one decimal place,the volume increases by a factor of $1.6$.

(C) Volume,$V = 0.03 \ m^3$
Temperature,$T = 129 + 273 = 402 \ K$
Mass of methane,$w = 6.0 \ g$
Molar mass of methane,$M = 12.01 + 4 \times 1.01 = 16.05 \ g \ mol^{-1}$
Using the ideal gas equation,$pV = nRT$,where $n = \frac{w}{M}$
$p = \frac{wRT}{MV} = \frac{6.0 \times 8.314 \times 402}{16.05 \times 0.03}$
$p = \frac{20065.392}{0.4815} \approx 41672.67 \ Pa$
Rounding to the nearest provided option,the value is $41648 \ Pa$.

(D) At equilibrium,the water vapour exerts a pressure equal to its saturated vapour pressure.
Using the ideal gas equation,$PV = nRT$,where $P = 3170 \ Pa$,$V = 1.0 \ dm^3 = 1.0 \times 10^{-3} \ m^3$,$T = 300 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$n = \frac{PV}{RT} = \frac{3170 \times 1.0 \times 10^{-3}}{8.314 \times 300} \ mol$.
$n = \frac{3.17}{2494.2} \ mol \approx 1.27 \times 10^{-3} \ mol$.

(A) Initially,the number of moles of gas in each bulb is $n_1 = \frac{P_i V}{R T_1}$ and $n_2 = \frac{P_i V}{R T_1}$.
After the temperature of the second bulb is raised to $T_2$,the number of moles of gas in both bulbs are $n_1' = \frac{P_f V}{R T_1}$ and $n_2' = \frac{P_f V}{R T_2}$.
Since the total number of moles of gas in both bulbs remains constant:
$n_1 + n_2 = n_1' + n_2'$
$\frac{P_i V}{R T_1} + \frac{P_i V}{R T_1} = \frac{P_f V}{R T_1} + \frac{P_f V}{R T_2}$
$\frac{2 P_i V}{R T_1} = \frac{P_f V}{R} \left( \frac{1}{T_1} + \frac{1}{T_2} \right)$
$\frac{2 P_i}{T_1} = P_f \left( \frac{T_2 + T_1}{T_1 T_2} \right)$
$P_f = \frac{2 P_i T_2}{T_1 + T_2}$

(B) Using the ideal gas equation: $PM = dRT$
Given: $P = 1 \ atm$,$T = 273 \ K$,$d = 10 \ mg/mL = 10 \ g/L$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$
Substituting the values: $1 \times M = 10 \times 0.0821 \times 273$
$M = 224.133 \ g/mol \approx 224 \ g/mol$
Since the species is diatomic $(A_2)$,the molar mass $M = 2 \times (\text{atomic weight of } A)$
$2 \times A = 224$
$A = 112 \ g/mol$

(C) For an ideal gas,the equation is $PV = nRT$.
$A$. $P$ vs $\frac{1}{V}$ at constant $T$ & $n$: Since $P = (nRT) \times \frac{1}{V}$,this is a linear equation of the form $y = mx$ passing through the origin. It is a straight line.
$B$. $PV$ vs $V^2$ at constant $T$ & $n$: Since $PV = nRT$ (a constant),the graph of $PV$ vs $V^2$ is a horizontal line parallel to the $V^2$ axis. It is a straight line.
$C$. $P$ vs $T^2$ at constant $V$ & $n$: From $P = (\frac{nR}{V})T$,we see that $P \propto T$. Therefore,$P$ vs $T^2$ is a parabolic curve,not a straight line.
$D$. $log\ P$ vs $log\ (V^2)$ at constant $T$ & $n$: Since $PV = k$,$P = k V^{-1}$. Taking log on both sides,$log\ P = log\ k - log\ V$. Since $log\ (V^2) = 2 log\ V$,we have $log\ P = log\ k - \frac{1}{2} log\ (V^2)$. This is a linear equation of the form $y = mx + c$. It is a straight line.

(A) At $STP$,$P_1 = 750 \, torr$,$V_1 = 127.4 \, mL$,and $T_1 = 273 \, K$.
When collected over water,the pressure of the dry gas $(P_{dry})$ is the total pressure minus the aqueous tension: $P_2 = P_{total} - P_{H_2O} = 725 \, torr - 25 \, torr = 700 \, torr$.
The temperature is $T_2 = 27 + 273 = 300 \, K$.
Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Substituting the values: $\frac{750 \times 127.4}{273} = \frac{700 \times V_2}{300}$.
$V_2 = \frac{750 \times 127.4 \times 300}{273 \times 700} \approx 150 \, mL$.

(A) According to the ideal gas equation,$PV = nRT$,which can be rearranged as $V = (\frac{nR}{P})T$.
This represents a straight line equation $y = mx$,where the slope $m = \frac{nR}{P}$.
Since the slope is inversely proportional to pressure $(m \propto \frac{1}{P})$,a smaller slope corresponds to a higher pressure.
From the given graph,the slope of the line for $P_1$ is the smallest,followed by $P_3$,and then $P_2$ (which has the largest slope).
Therefore,the correct order of pressures is $P_1 > P_3 > P_2$.

(B) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Given:
$P_1 = P$
$V_1 = 100\, \text{cc}$
$T_1 = 0\, ^\circ C = 273\, \text{K}$
New conditions:
$P_2 = 1.5 P = \frac{3}{2} P$
$T_2 = T_1 + \frac{1}{3} T_1 = \frac{4}{3} T_1 = \frac{4}{3} \times 273\, \text{K}$
Substituting these values into the formula:
$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{P \times 100 \times (\frac{4}{3} \times 273)}{(\frac{3}{2} P) \times 273}$
$V_2 = 100 \times \frac{4}{3} \times \frac{2}{3} = \frac{800}{9} \approx 88.88\, \text{cc}$
Rounding to one decimal place,$V_2 = 88.9\, \text{cc}$.

(A) From the ideal gas equation,$PV = nRT = \frac{w}{M}RT$,we have $P = \frac{dRT}{M}$,where $d$ is the density,$R$ is the gas constant,$T$ is the temperature,and $M$ is the molar mass.
For the same gas,$M$ is constant,so $P \propto dT$.
Therefore,the ratio of pressures is $\frac{P_1}{P_2} = \frac{d_1}{d_2} \times \frac{T_1}{T_2}$.
Given $\frac{d_1}{d_2} = \frac{1}{2}$ and $\frac{T_1}{T_2} = \frac{2}{1}$.
Substituting these values,$\frac{P_1}{P_2} = \frac{1}{2} \times \frac{2}{1} = \frac{1}{1}$.
Thus,the ratio of their respective pressures is $1:1$.

(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{w}{M}$.
Given:
Mass of methane $(w)$ = $6.0 \ g$
Molar mass of methane $(CH_4)$ $(M)$ = $12.01 + 4 \times 1.01 = 16.05 \ g/mol$
Volume $(V)$ = $0.03 \ m^3$
Temperature $(T)$ = $129 + 273.15 = 402.15 \ K$ (approximated to $402 \ K$ for calculation)
Gas constant $(R)$ = $8.314 \ J K^{-1} mol^{-1}$
Substituting the values into the equation:
$P \times 0.03 = \frac{6.0}{16.05} \times 8.314 \times 402.15$
$P \times 0.03 = 0.37383 \times 8.314 \times 402.15$
$P \times 0.03 = 1249.53$
$P = \frac{1249.53}{0.03} \approx 41651 \ Pa$
Rounding to the nearest provided option,the correct answer is $41648 \ Pa$.

(D) From the ideal gas equation,$PV = nRT = (\frac{m}{M})RT$.
Rearranging this,$P(\frac{m}{V}) = \frac{PRT}{M}$,which gives $Pd = \frac{PM}{RT}$.
For a given temperature $T$,the slope of the $Pd$ vs $P$ graph is $\frac{M}{RT}$,which is directly proportional to the molar mass $M$ of the gas.
The molar masses are: $CO_2 = 44\,g/mol$,$N_2 = 28\,g/mol$,$He = 4\,g/mol$,$H_2 = 2\,g/mol$.
Since the slope increases as molar mass increases,the order of slopes is $CO_2 > N_2 > He > H_2$.
Looking at the graph,the slope follows the order $1 > 2 > 3 > 4$.
Therefore,$1-CO_2, 2-N_2, 3-He, 4-H_2$.
Matching this with the given options,option $D$ is $2-N_2, 3-He, 1-CO_2, 4-H_2$.

(B) The total number of moles of $H_2$ gas remains constant when the valve is opened.
$n_{total} = n_A + n_B$
Using the ideal gas equation $PV = nRT$,we have $n = \frac{PV}{RT}$.
$n_A = \frac{4 \ atm \times 2 \ L}{R \times 400 \ K} = \frac{8}{400R} = \frac{0.02}{R}$
$n_B = \frac{8 \ atm \times 4 \ L}{R \times 800 \ K} = \frac{32}{800R} = \frac{0.04}{R}$
$n_{total} = \frac{0.02}{R} + \frac{0.04}{R} = \frac{0.06}{R}$
After opening the valve,the gas occupies both compartments. Since the vessels are thermally insulated,the temperature in each compartment remains at its initial value ($400 \ K$ and $800 \ K$ respectively).
$n_{total} = \frac{P_f V_A}{R T_A} + \frac{P_f V_B}{R T_B}$
$\frac{0.06}{R} = P_f \left[ \frac{2}{400R} + \frac{4}{800R} \right]$
$0.06 = P_f \left[ 0.005 + 0.005 \right]$
$0.06 = P_f \times 0.01$
$P_f = 6 \ atm$

(A) For container $X$,the initial state is $P_1 V = n_1 RT$. After expansion to $2V$ with pressure change $\Delta P$,the final state is $(P_1 - \Delta P) 2V = n_1 RT$.
Equating the two: $P_1 V = 2P_1 V - 2\Delta P V$,which gives $P_1 V = 2\Delta P V$.
Thus,$n_1 = \frac{2\Delta P V}{RT}$.
For container $Y$,the initial state is $P_2 V = n_2 RT$. After expansion to $2V$ with pressure change $0.2 \Delta P$,the final state is $(P_2 - 0.2 \Delta P) 2V = n_2 RT$.
Equating the two: $P_2 V = 2P_2 V - 0.4 \Delta P V$,which gives $P_2 V = 0.4 \Delta P V$.
Thus,$n_2 = \frac{0.4 \Delta P V}{RT}$.
The ratio of masses is equal to the ratio of moles: $\frac{w_X}{w_Y} = \frac{n_1}{n_2} = \frac{2\Delta P V / RT}{0.4 \Delta P V / RT} = \frac{2}{0.4} = 5$.

(B) According to Charles's law,$V \propto T$ at constant pressure. Since density $d = \frac{m}{V}$,we have $d \propto \frac{1}{T}$,which implies $d_1 T_1 = d_2 T_2$.
Let the absolute zero be $x\,^o C$. Then $T_1 = (25 - x) \text{ K}$ and $T_2 = (100 - x) \text{ K}$.
Given $d_1 = 1.2\ gL^{-1}$ and $d_2 = 0.9\ gL^{-1}$.
Substituting the values: $1.2 \times (25 - x) = 0.9 \times (100 - x)$.
$1.2(25) - 1.2x = 0.9(100) - 0.9x$.
$30 - 1.2x = 90 - 0.9x$.
$30 - 90 = 1.2x - 0.9x$.
$-60 = 0.3x$.
$x = \frac{-60}{0.3} = -200\,^o C$.

(C) For an adiabatic process, $TV^{\gamma-1} = \text{constant}$.
Taking log on both sides: $log \, T + (\gamma-1) log \, V = \text{constant}$.
This is the equation of a straight line $y = mx + c$, where $y = log \, T$, $x = log \, V$, and the slope $m = -(\gamma-1)$.
From the graph, the slope $m = \frac{1.40 - 1.60}{1.60 - 1.10} = \frac{-0.20}{0.50} = -0.4$.
Equating the slopes: $-(\gamma-1) = -0.4$, so $\gamma-1 = 0.4$, which gives $\gamma = 1.4$.
For a diatomic gas, $\gamma = 1.4$ (e.g., $C_p/C_v = 7/5 = 1.4$).
Thus, the gas is diatomic and undergoing an adiabatic change.

(D) $1$. Calculate the volume of the alveolus $(V)$: $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (0.0050 \ cm)^3 = 5.236 \times 10^{-7} \ cm^3 = 5.236 \times 10^{-13} \ m^3 = 5.236 \times 10^{-10} \ L$.
$2$. Use the ideal gas law $(PV = nRT)$ to find the total moles of air $(n_{air})$: $n_{air} = \frac{PV}{RT} = \frac{1.0 \ atm \times 5.236 \times 10^{-10} \ L}{0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 310 \ K} = 2.057 \times 10^{-11} \ mol$.
$3$. Calculate the moles of oxygen $(n_{O_2})$: $n_{O_2} = 0.14 \times n_{air} = 0.14 \times 2.057 \times 10^{-11} = 2.88 \times 10^{-12} \ mol$.
$4$. Calculate the number of oxygen molecules $(N)$: $N = n_{O_2} \times N_A = 2.88 \times 10^{-12} \times 6.022 \times 10^{23} \approx 1.73 \times 10^{12}$ molecules.

(C) Since the number of moles of gas remains constant,we use the combined gas law: $\frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f}$.
Given:
$P_i = 3.21 \times 10^5 \ Pa$
$T_i = -5.0 \ ^\circ C = 268.15 \ K$
$T_f = 28 \ ^\circ C = 301.15 \ K$
$V_f = V_i + 0.03 V_i = 1.03 V_i$
Rearranging for $P_f$:
$P_f = P_i \times \frac{V_i}{V_f} \times \frac{T_f}{T_i}$
Substituting the values:
$P_f = 3.21 \times 10^5 \ Pa \times \frac{1}{1.03} \times \frac{301.15}{268.15}$
$P_f \approx 3.50 \times 10^5 \ Pa$.

(C) According to the kinetic molecular theory of gases,the assumptions that the volume of gas molecules is negligible compared to the total volume of the gas and that there are no intermolecular forces of attraction are valid for an ideal gas.
These conditions are best approximated by real gases at $high \ temperature$ and $low \ pressure$.
At $high \ temperature$,the kinetic energy of the molecules is high,which overcomes the intermolecular forces of attraction.
At $low \ pressure$,the molecules are far apart,making the volume occupied by the molecules themselves negligible compared to the total volume of the container.

(B) According to the ideal gas equation,$PV = nRT$,which can be rearranged as $P = (nRT) \cdot (1/V)$.
For a fixed amount of gas ($n$ is constant),the product $PV$ is directly proportional to the temperature $T$.
In a $P$ versus $V$ graph (isotherm),for a given volume $V$,the pressure $P$ is higher for higher temperatures.
Looking at the graph,at any constant volume $V$,the pressure values follow the order $P_1 > P_2 > P_3$ for the curves corresponding to $T_1, T_2, T_3$ respectively.
Therefore,the relationship between the temperatures is $T_1 > T_2 > T_3$.

(A) According to Charles's Law,for a fixed mass of gas at constant pressure,$V \propto T$ (where $T$ is in Kelvin).
Initial conditions: $V_1 = 0.4 \, L$,$T_1 = 0 + 273 = 273 \, K$.
Final conditions: $V_2 = ?$,$T_2 = 273 + 273 = 546 \, K$.
Using the formula $\frac{V_1}{T_1} = \frac{V_2}{T_2}$:
$\frac{0.4}{273} = \frac{V_2}{546}$
$V_2 = 0.4 \times \frac{546}{273} = 0.4 \times 2 = 0.8 \, L$.

(C) According to Charles's Law,at constant pressure,$V \propto T$.
Thus,$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$.
Given: $V_{1} = 200 \, cm^{3}$,$T_{1} = 17 + 273 = 290 \, K$,$T_{2} = 47 + 273 = 320 \, K$.
Substituting the values: $\frac{200}{290} = \frac{V_{2}}{320}$.
$V_{2} = \frac{200 \times 320}{290} = 220.68 \, cm^{3} \approx 220.6 \, cm^{3}$.

(B) The ideal gas equation is $PV = nRT$,which can be written as $P = \frac{dRT}{M}$,where $d$ is density and $M$ is molecular mass.
Given: $d = 0.00356 \ g/mL = 3.56 \ g/L$,$P = 1 \ atm$,$R = 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 273 \ K$.
Rearranging for $M$: $M = \frac{dRT}{P}$.
Substituting the values: $M = \frac{3.56 \times 0.0821 \times 273}{1} \approx 79.75 \approx 80 \ g/mol$.

(B) Step $1$: Calculate the moles of each gas.
Moles of $N_2 = \frac{3.01 \times 10^{24}}{6.02 \times 10^{23}} = 5 \ mol$.
Moles of $O_2 = \frac{32 \ g}{32 \ g/mol} = 1 \ mol$.
Total moles $(n)$ = $5 + 1 = 6 \ mol$.
Step $2$: Calculate the total volume of the mixture using the ideal gas equation $PV = nRT$.
$V = \frac{nRT}{P} = \frac{6 \times (1/12) \times 860}{3}$.
$V = \frac{0.5 \times 860}{3} = \frac{430}{3} \approx 143.33 \ L$.
Step $3$: Calculate the total mass of the mixture.
Mass of $N_2 = 5 \ mol \times 28 \ g/mol = 140 \ g$.
Mass of $O_2 = 32 \ g$.
Total mass = $140 + 32 = 172 \ g$.
Step $4$: Calculate the density $(d = \frac{mass}{volume})$.
$d = \frac{172}{143.33} \approx 1.2 \ g/L$.

(B) Using the ideal gas equation: $PV = nRT = \frac{m}{M}RT$,where $M$ is the molar mass and $d = \frac{m}{V}$ is the density.
The formula for molar mass is $M = \frac{dRT}{P}$.
Given: $d = 3.84 \ g/L$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,$T = 210 + 273 = 483 \ K$,and $P = 1 \ atm$.
Substituting the values: $M = \frac{3.84 \times 0.0821 \times 483}{1} \approx 152.2 \ g/mol$.
Calculating the molar mass for the options:
$A: C_{10}H_{14}O = (10 \times 12) + (14 \times 1) + 16 = 150 \ g/mol$
$B: C_{10}H_{16}O = (10 \times 12) + (16 \times 1) + 16 = 152 \ g/mol$
$C: C_{10}H_{16}O_2 = (10 \times 12) + (16 \times 1) + 32 = 168 \ g/mol$
$D: C_{10}H_{18}O = (10 \times 12) + (18 \times 1) + 16 = 154 \ g/mol$
Since the calculated molar mass is $152 \ g/mol$,the correct formula is $C_{10}H_{16}O$.

(C) From the ideal gas equation,$PV = nRT$.
Since $n = \frac{m}{M}$,we have $PV = \frac{mRT}{M}$.
Rearranging for density $(d = \frac{m}{V})$,we get $P = \frac{dRT}{M}$,which implies $d = \frac{PM}{RT}$.
At constant temperature and pressure,$d \propto M$.
Therefore,the ratio of densities is $\frac{d_{NH_3}}{d_{HCl}} = \frac{M_{NH_3}}{M_{HCl}}$.
The molar mass of $NH_3 = 14 + 3(1) = 17 \ g/mol$.
The molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
Ratio $= \frac{17}{36.5} \approx 0.4657 \approx 0.46$.

(C) The density $(\rho)$ of a gas is given by the formula $\rho = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature.
Given that the temperature $T$ is constant for both gases.
For $N_2$ gas: $\rho_{N_2} = \frac{P_{N_2} \times M_{N_2}}{RT} = \frac{4 \times 28}{RT}$.
For the unknown gas: $\rho_{gas} = \frac{P_{gas} \times M_{gas}}{RT} = \frac{2 \times M_{gas}}{RT}$.
According to the problem,$\rho_{gas} = 2 \times \rho_{N_2}$.
Substituting the expressions: $\frac{2 \times M_{gas}}{RT} = 2 \times \left( \frac{4 \times 28}{RT} \right)$.
Canceling $RT$ from both sides: $2 \times M_{gas} = 8 \times 28$.
$M_{gas} = 4 \times 28 = 112\, g\, mol^{-1}$.

(A) According to Boyle's law,for a fixed amount of gas at constant temperature,$P_1V_1 = P_2V_2$.
Given: $V_1 = 750.0 \ mL$,$P_1 = 840.0 \ mm \ Hg$,$P_2 = 360.0 \ mm \ Hg$.
Substituting the values: $840.0 \ mm \ Hg \times 750.0 \ mL = 360.0 \ mm \ Hg \times V_2$.
$V_2 = \frac{840.0 \times 750.0}{360.0} \ mL = 1750 \ mL$.
Converting to liters: $1750 \ mL = 1.750 \ L$.

(B) According to Boyle's law,$PV = \text{constant}$.
Taking logarithm on both sides:
$\log \ P + \log \ V = \log \ (\text{constant})$
$\log \ P = - \log \ V + \log \ (\text{constant})$
This equation is of the form $y = mx + c$,where $y = \log \ P$,$x = \log \ V$,and the slope $m = -1$.
Hence,the plot of $\log \ P$ vs $\log \ V$ is a straight line with a negative slope.

(C) Let the volume of the vessel be $V$. Initially,the volume of air at $T_1 = 300 \ K$ is $V$.
Since $2/5$ of the air is expelled,the remaining volume of air at $300 \ K$ is $V - (2/5)V = (3/5)V$.
When the vessel is heated to a new temperature $T_2$,this remaining air expands to fill the entire volume $V$ of the vessel.
Since the pressure remains constant in an open vessel,we use Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Substituting the values: $\frac{(3/5)V}{300} = \frac{V}{T_2}$.
Solving for $T_2$: $T_2 = 300 \times \frac{5}{3} = 500 \ K$.

(D) Using the ideal gas equation: $PV = nRT$
Given: $P = 200 \, Pa$,$V = 10 \, m^3$,$T = 1000 \, K$,$n = (0.5 + x) \, mol$.
Substituting the values: $200 \times 10 = (0.5 + x) \times R \times 1000$
$2000 = (0.5 + x) \times 1000R$
$2 = (0.5 + x)R$
$\frac{2}{R} = 0.5 + x$
$x = \frac{2}{R} - 0.5$
$x = \frac{2 - 0.5R}{R} = \frac{4 - R}{2R}$

(C) For an open vessel,the pressure $P$ and volume $V$ remain constant. According to the ideal gas law,$PV = nRT$,which implies $n_1T_1 = n_2T_2$.
Let the initial number of moles be $n_1 = n$ and the initial temperature $T_1 = 27 + 273 = 300\,K$.
Since two-fifths of the air escapes,the remaining moles are $n_2 = n - \frac{2}{5}n = \frac{3}{5}n$.
Substituting these values into the equation: $n \times 300 = \frac{3}{5}n \times T_2$.
Solving for $T_2$: $T_2 = 300 \times \frac{5}{3} = 500\,K$.

(A) The density $d$ of a gas is given by the formula $d = \frac{PM}{RT}$.
Here,$P = \frac{800}{760}\, atm$,$M = 44\, g/mol$ (molar mass of $CO_2$),$R = 0.0821\, L\, atm\, K^{-1} mol^{-1}$,and $T = 100 + 273 = 373\, K$.
Substituting the values: $d = \frac{(\frac{800}{760}) \times 44}{0.0821 \times 373} \approx 1.51\, g/L$.

(A) For an ideal gas at constant pressure,the work done $W$ is given by the formula $W = P \Delta V$.
According to the ideal gas equation,$PV = nRT$,so at constant pressure,$P \Delta V = nR \Delta T$.
Given:
$n = 1 \, mol$
$\Delta T = 2 \, ^oC = 2 \, K$ (since the change in temperature is the same in Celsius and Kelvin scales).
Substituting the values:
$W = nR \Delta T = 1 \, mol \times R \times 2 \, K = 2R$.
Since the gas expands when heated at constant pressure,the work is done by the system,which is conventionally represented as negative in some sign conventions,but in standard thermodynamic problems,the magnitude $2R$ is the expected answer. Given the options provided,the magnitude is $2R$.

(B) The ideal gas equation is $PV = nRT = \frac{w}{M} RT$.
Density,$\rho = \frac{w}{V} = \frac{PM}{RT}$.
Thus,the density of the ideal gas is directly proportional to the pressure $(P)$ and inversely proportional to the temperature $(T)$.
To maximize density,we require the highest pressure and the lowest temperature.
Comparing the given options:
$A$: $1 \, atm, 273 \, K$
$B$: $2 \, atm, 273 \, K$
$C$: $1 \, atm, 546 \, K$
$D$: $2 \, atm, 546 \, K$
Option $B$ provides the highest pressure $(2 \, atm)$ and the lowest temperature $(273 \, K)$,resulting in the maximum density.

(A) From the ideal gas equation,density $d = \frac{PM_w}{RT}$,where $P$ is pressure,$M_w$ is molar mass,$R$ is the gas constant,and $T$ is temperature.
Assuming the gases have the same molar mass $(M_w)$,the relationship is $d \propto \frac{P}{T}$ or $P \propto d \times T$.
Given the ratio of densities $\frac{d_1}{d_2} = \frac{1}{2}$ and the ratio of temperatures $\frac{T_1}{T_2} = \frac{2}{1}$.
The ratio of pressures is $\frac{P_1}{P_2} = \frac{d_1}{d_2} \times \frac{T_1}{T_2}$.
Substituting the values: $\frac{P_1}{P_2} = \frac{1}{2} \times \frac{2}{1} = \frac{1}{1}$.
Thus,the ratio of their pressures is $1:1$.

(D) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is mass and $M$ is molar mass.
Density $d = \frac{m}{V} = \frac{PM}{RT}$.
Since $P$ and $M$ are constant,$d \propto \frac{1}{T}$.
Given $T_1 = 27 + 273 = 300\, K$ and $d_1 = d$.
We need $T_2$ such that $d_2 = 0.75\, d$.
Using the relation $\frac{d_2}{d_1} = \frac{T_1}{T_2}$:
$\frac{0.75\, d}{d} = \frac{300}{T_2}$.
$0.75 = \frac{300}{T_2} \implies T_2 = \frac{300}{0.75} = 400\, K$.
Converting to Celsius: $400 - 273 = 127\,^oC$. Since $400\, K$ is provided as an option,the correct choice is $D$.

(D) The density of a gas is given by the formula $d = \frac{PM_w}{RT}$.
For the same gas,the ratio of densities at two different conditions is given by $\frac{d_2}{d_1} = \frac{P_2}{P_1} \times \frac{T_1}{T_2}$.
Given: $d_1 = 5.46\, g/L$,$P_1 = 2\, atm$,$T_1 = 27 + 273 = 300\, K$.
At $STP$: $P_2 = 1\, atm$,$T_2 = 273\, K$.
Substituting the values: $\frac{d_2}{5.46} = \frac{1}{2} \times \frac{300}{273}$.
$d_2 = 5.46 \times 0.5 \times 1.0989 \approx 3\, g/L$.

(D) According to Charles's Law,at constant pressure,$V \propto T$.
Therefore,$\frac{V_2}{V_1} = \frac{T_2}{T_1}$.
Given that the volume increases by $10 \%$,$V_2 = 1.10 V_1$.
Substituting this into the equation: $\frac{T_2}{T_1} = \frac{1.10 V_1}{V_1} = 1.10$.
The percentage increase in temperature is given by $\frac{T_2 - T_1}{T_1} \times 100 = (\frac{T_2}{T_1} - 1) \times 100$.
$= (1.10 - 1) \times 100 = 0.10 \times 100 = 10 \%$.

(A) According to Charles's Law,the volume $V_t$ at temperature $t\,^{\circ}C$ is given by:
$V_t = V_0 + \frac{V_0}{273}t$
This equation is in the form of a straight line $y = mx + c$,where $y = V_t$,$x = t$,$c = V_0$,and the slope $m = \frac{V_0}{273}$.
Therefore,the slope of the graph is $\frac{V_0}{273}$.

(C) According to Charles's Law,at constant pressure,$V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $T_1 = 0\,^\circ C = 273\, K$,$V_1 = V$,$V_2 = 2V$.
Substituting the values: $\frac{V}{273} = \frac{2V}{T_2}$.
$T_2 = 273 \times 2 = 546\, K$.
To convert to Celsius: $T(^{\circ}C) = 546 - 273 = 273\,^{\circ}C$.

(B) According to Charles's Law,at constant pressure,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 20 \ L$,$T_1 = 100 \ K$,$T_2 = 300 \ K$.
Substituting the values: $\frac{20}{100} = \frac{V_2}{300}$.
$V_2 = \frac{20 \times 300}{100} = 60 \ L$.
The change in volume $\Delta V = V_2 - V_1 = 60 \ L - 20 \ L = 40 \ L$.

(C) Using the ideal gas equation $PV = \frac{W R T}{M}$,where $V, M$,and $R$ are constant for the same container and same gas:
$\frac{P_1}{P_2} = \frac{W_1 T_1}{W_2 T_2}$
Substituting the given values:
$\frac{1}{10} = \frac{3.2 \times 200}{W_2 \times 500}$
$W_2 = \frac{3.2 \times 200 \times 10}{500}$
$W_2 = \frac{3.2 \times 20}{5} = 3.2 \times 4 = 12.8 \, g$

(D) According to Charles's Law,for a fixed mass of gas at constant pressure,$V \propto T$,which can be written as $V = (nR/P)T$.
Comparing this with the equation of a straight line $y = mx$,where $y = V$ and $x = T$,the slope $m = nR/P$.
Since the slope is inversely proportional to pressure $(m \propto 1/P)$,the line with the smallest slope corresponds to the highest pressure.
From the given graph,the slope of the line for $P_3$ is the smallest,followed by $P_2$,and the slope for $P_1$ is the largest.
Therefore,the relationship between the pressures is $P_3 > P_2 > P_1$.